# Integrate $\int \ln(x^2 +1)\ dx$

$$\int \ln(x^2 +1)\ dx$$

I done it using integration by parts where

$\int u\ dv = uv - \int v\ du$

Let $u$ = $\ln(x^2 +1)$

$du = \frac{2x}{x^2+1} dx$

Let $dv = dx$ so $v=x$

$\int \ln(x^2 +1)\ dx = x \ln (x^2 +1) - \int \frac{2x^2}{x^2+1}$

I integrate $2\int \frac{x^2}{x^2+1}$ separately using substitution.

Let $u$ = $x^2 + 1$ -> $dx= \frac{1}{2x} du$

substitute $u$ back into it I get a simplified $\int \frac{x}{x^2 + 1} dx$

I used substitution again and got $\frac{1}{2} \int \frac{1}{u} du$

When I sub it back into the original question, this is my final answer,

$x\ln(x^2 +1) + \frac{1}{2} \ln(x^2 +1) + C$

The answer is, $x\ln(x^2 +1) - 2x + 2\tan^{-1} (x) + C$

I believe my integration of $2\int \frac{x^2}{x^2+1}$ is wrong. Where did I went wrong ?

Your integration by parts in the beginning is correct. Truly :

$${\displaystyle\int}\ln\left(x^2+1\right)\,\mathrm{d}x =x\ln\left(x^2+1\right)-{\displaystyle\int}\dfrac{2x^2}{x^2+1}\,\mathrm{d}x$$

Now, handling the second integral, first of all let's factor out the constant and write $x^2$ as $x^2 + 1 - 1$ to split it up :

$${\displaystyle\int}\dfrac{2x^2}{x^2+1}\,\mathrm{d}x =\class{steps-node}{\cssId{steps-node-1}{2}}{\displaystyle\int}\dfrac{x^2}{x^2+1}\,\mathrm{d}x ={\displaystyle\int}\left(\dfrac{\class{steps-node}{\cssId{steps-node-4}{x^2+1}}}{x^2+1}-\dfrac{\class{steps-node}{\cssId{steps-node-5}{1}}}{x^2+1}\right)\mathrm{d}x$$

$$=$$

$$={\displaystyle\int}\left(1-\dfrac{1}{x^2+1}\right)\mathrm{d}x ={\displaystyle\int}1\,\mathrm{d}x-{\displaystyle\int}\dfrac{1}{x^2+1}\,\mathrm{d}x =2x-2\arctan\left(x\right)$$

The integral is now solved and we yield the result :

$${\displaystyle\int}\ln\left(x^2+1\right)\,\mathrm{d}x=x\ln\left(x^2+1\right)+2\arctan\left(x\right)-2x + C$$

• Why can’t I use substitution twice to solve the second integral Commented Apr 22, 2018 at 13:28
• @user59439 The expression you yield via the other substitution is mistaken. You replaced the differential by the one you found but the expression remained unchanged. You still need to express the $x$ parts in therms of the $u$ you substituted. This is why you got a mixed result. Commented Apr 22, 2018 at 13:33

$2\int \frac{x^2}{x^2+1}$

Does not equal the expression with just an x in the numerator after the substitution . You can add subtract 1 in numerator to integrate $2\int \frac{x^2}{x^2+1}$

This is wrong :- $2\int \frac{x^2}{x^2+1}$ This is what you get after applying by-parts :- $2\int \frac{x^2}{x^2+1} dx$ Since differential of x is $dx$ . In by parts formula we have differential of $u$ not it's differenciation.

• The $\mathrm{d}x$ is not important as long as it's clear what you are intagrating with respect to. Commented Apr 22, 2018 at 14:20